This is regarding Theorem 1.8 on page 11 of Rudin's Real and Complex Analysis, we have a measurable space $X$ and two real valued functions on it $u$ and $v$. We then have a continuous mapping $\Phi$ from the real plane into a topological space $Y$. We define $f(x)=(u(x),y(x))$ and $h=\Phi\circ f$, and from here we need to show that $h$ is measurable and we know that it suffices to show that $f$ is measurable.
We then take a rectangle $R$ in the plane with sides parallel to the axes, which we write as the Cartesian product of $I_1$ and $I_2$ which are two segments, presumably two adjacent sides of the rectangle. We then say that: $$f^{-1}(R)=u^{-1}(I_1)\cap v^{-1}(I_2) \tag{1},$$ and this is what's confusing me. $I_1$ and $I_2$ are essentially just lines in the plane, why would the intersection of their preimage be equal to the preimage of the entire rectangle and not just two "lines" in $X$?
Disclaimer: I am a physicist not a mathematician please have mercy on my soul if this is a trivial error on my part.
As far as I understand, we have $R=I_1\times I_2$.
Now we want to show $f^{-1}(R)=u^{-1}(I_1)\cap v^{-1}(I_2)$.
This is a straight forward proof:
$f^{-1}(R)=\{x\in X: f(x)\in R\}=\{x\in X: (u(x),v(x))\in I_1\times I_2\}=\{x\in X: u(x)\in I_1~\text{and}~ v(x)\in I_2\}$
$=\{x\in X: u(x)\in I_1\}\cap \{x\in X: v(x)\in I_2\}=u^{-1}(I_1)\cap v^{-1}(I_2)$